假設(shè)您在Linux上運(yùn)行Django,并且您有一個視圖,并且您希望該視圖從名為cmd的子進(jìn)程返回數(shù)據(jù),該子進(jìn)程對視圖創(chuàng)建的文件進(jìn)行操作,例如：likeo：

 def call_subprocess(request):
     response = HttpResponse()

     with tempfile.NamedTemporaryFile("W") as f:
         f.write(request.GET['data']) # i.e. some data

     # cmd operates on fname and returns output
     p = subprocess.Popen(["cmd", f.name], 
                   stdout=subprocess.PIPE, 
                   stderr=subprocess.PIPE)

     out, err = p.communicate()

     response.write(p.out) # would be text/plain...
     return response

現(xiàn)在,假設(shè)cmd的啟動時間非常慢,但運(yùn)行時間非?？?并且它本身沒有守護(hù)進(jìn)程模式.我想改善這種觀點(diǎn)的響應(yīng)時間.

我想通過在工作池中啟動一些cmd實(shí)例,讓它們等待輸入,并讓call_process請求其中一個工作池進(jìn)程處理數(shù)據(jù),使整個系統(tǒng)運(yùn)行得更快.

這實(shí)際上是兩個部分：

第1部分.調(diào)用cmd和cmd的函數(shù)等待輸入.這可以通過管道完成,即

def _run_subcmd():
    p = subprocess.Popen(["cmd", fname], 
        stdout=subprocess.PIPE, stderr=subprocess.PIPE)

    out, err = p.communicate()
    # write 'out' to a tmp file
    o = open("out.txt", "W")
    o.write(out)
    o.close()
    p.close()
    exit()

def _run_cmd(data):
    f = tempfile.NamedTemporaryFile("W")
    pipe = os.mkfifo(f.name)

    if os.fork() == 0:
        _run_subcmd(fname)
    else:
        f.write(data)

    r = open("out.txt", "r")
    out = r.read()
    # read 'out' from a tmp file
    return out

def call_process(request):
    response = HttpResponse()

    out = _run_cmd(request.GET['data'])

    response.write(out) # would be text/plain...
    return response

第2部分.在后臺運(yùn)行的一組正在等待數(shù)據(jù)的工作者.即,我們希望擴(kuò)展上述內(nèi)容,以便子進(jìn)程已經(jīng)運(yùn)行,例如當(dāng)Django實(shí)例初始化時,或者首次調(diào)用此call_process時,會創(chuàng)建一組這些worker

WORKER_COUNT = 6
WORKERS = []

class Worker(object):
    def __init__(index):
        self.tmp_file = tempfile.NamedTemporaryFile("W") # get a tmp file name
        os.mkfifo(self.tmp_file.name)
        self.p = subprocess.Popen(["cmd", self.tmp_file], 
            stdout=subprocess.PIPE, stderr=subprocess.PIPE)
        self.index = index

    def run(out_filename, data):
        WORKERS[self.index] = Null # qua-mutex??
        self.tmp_file.write(data)
        if (os.fork() == 0): # does the child have access to self.p??
            out, err = self.p.communicate()
            o = open(out_filename, "w")
            o.write(out)
            exit()

        self.p.close()
        self.o.close()
        self.tmp_file.close()
        WORKERS[self.index] = Worker(index) # replace this one
        return out_file

    @classmethod
    def get_worker() # get the next worker
    # ... static, incrementing index 

應(yīng)該在某處對工作人員進(jìn)行一些初始化,如下所示：

def init_workers(): # create WORKERS_COUNT workers
    for i in xrange(0, WORKERS_COUNT):
        tmp_file = tempfile.NamedTemporaryFile()
        WORKERS.push(Worker(i))

現(xiàn)在,我上面的內(nèi)容變成了以下內(nèi)容：

def _run_cmd(data):
     Worker.get_worker() # this needs to be atomic & lock worker at Worker.index

     fifo = open(tempfile.NamedTemporaryFile("r")) # this stores output of cmd

     Worker.run(fifo.name, data)
     # please ignore the fact that everything will be
     # appended to out.txt ... these will be tmp files, too, but named elsewhere.

     out = fifo.read()
     # read 'out' from a tmp file
     return out


def call_process(request):
     response = HttpResponse()

     out = _run_cmd(request.GET['data'])

     response.write(out) # would be text/plain...
     return response

現(xiàn)在,問題：

>這會有用嗎？ (我只是把它從頭頂輸入StackOverflow,所以我確定存在問題,但從概念上講,它會起作用)
>要尋找的問題是什么？
>有更好的替代品嗎？例如線程是否也能正常運(yùn)行(它是Debian Lenny Linux)？是否有任何庫可以處理這樣的并行流程工作池？
>我應(yīng)該注意與Django的交互嗎？

謝謝閱讀！我希望你覺得這和我一樣有趣.

布賴恩

解決方法:

看起來我正在推銷這款產(chǎn)品,因?yàn)檫@是我第二次回復(fù)推薦.

但似乎您需要一個Message Queing服務(wù),特別是一個分布式消息隊(duì)列.

它是如何工作的：

>您的Django應(yīng)用程序請求CMD
> CMD被添加到隊(duì)列中
> CMD被推到了幾個作品
>它被執(zhí)行并且結(jié)果返回上游

大多數(shù)代碼都存在,您不必去構(gòu)建自己的系統(tǒng).

看看最初用Django構(gòu)建的Celery.

http://www./
http:///blog/2009/09/10/rabbitmq-celery-and-django/